Asubmarine has a "crush depth" (that is, the depth at which water pressure
will crush the submarine) of 250 m. what is the approximate pressure (water
plus atmospheric) at this depth? (recall that the density of seawater is 1025
kg/m3 g = 9.81 m/s2, and 1 kg/(m. s) = 1 pa = 9.8692 x 10-6 atm.)

25.8 atm

Explanation:

Total pressure is:

P = Patm + ρgh

where Patm is the atmosphere pressure,

ρ is the fluid density,

g is acceleration due to gravity,

and h is the depth of the fluid.

P = (1 atm) + (1025 kg/m³) (9.81 m/s²) (250 m) (9.8692×10⁻⁶ atm/Pa)

P = 1 atm + 24.8 atm

P = 25.8 atm

v=ir = 3x15 =45 v

there is no correct list among the choices you've provided.

the list might look something like this:

-- baby human

-- piccolo

-- flute

-- soprano singer

-- trumpet

-- tenor singer

-- trombone

-- baritone singer

-- baritone horn

-- bass singer

-- tuba

since i had to make up the list myself, i made it easy on myself, and i didn't try to shuffle the string instruments and woodwinds into it too.